Probability Density Under Transformation

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1 Probability Density Under Transformation Pramook Khungurn September 5, 5 Introduction In creating an algorithm that samples points from some domain, a problem that always comes up is the following: Let A and B be sets, p A ( ) be a probability density on A, and f be a function from A to B. If one samples x from A according to p A, then what is the probability density of f(x)? This document discusses the solution to the above problem and its application to construction of sampling algorithm. One-Dimensional Case. The Main Theorem We first start with the simplest case where A and B are both subsets of the real line R. Let x A. The number p A (x) means that, in the infinitesimal interval [x, x + δx), there exists p A (x)δx amount of probability mass. Here, δx is a differential quantity such that (δx) =. Assume that f is continuous and infinitely differentiable. The function f sends the interval [x, x + δx) to the interval [f(x), f(x + δx)). By Taylor expansion, f(x + δx) = f(x) + f (x)δx + O((δx) ) = f(x) + f (x)δx. So, the resuling interval is [f(x), f(x) + f (x)δx), which as width f (x) δx. This means that the mass p A (x)δx gets distributed to an interval of width f (x)δx. As a result: Density at point f(x) = p A(x)δx f (x) δx = p A(x) f (x). This density is defined only when f (x), which means that f is one-to-one in a neighborhood of x. As such, we have the following theorem. Theorem. Let A and B be subsets of R, p A be a probability density on A, f : A B be continuous and differentiable and f (x) for all x A. The induced probability density p B ( ) arisen from the process of sampling x according to p A and then computing f(x) is given by: p B (f(x)) = p A(x) f (x).

2 . The Inversion Method The above theorem can be used to create sampling algorithm for any integrable density function on the real line from a uniformly random sample from the interval [, ). In this situation, A = [, ) and p A (x) = for all x A. The density p B ( ) is given to us. We want to find function f : A B such that, for any x A: Multiply both sides by f (x), we have: Let P B be the CDF of p B : We have that: p B (f(x)) = p A(x) f (x) = f (x). p B (f(x)) f (x) =. P B (y) = y p B (t) dt. {P B (f(x))} = p B (f(x))f (x) = p B (f(x)) f (x) given that f is an increasing function. Let us assume that f is increasing for now. We have that Integrating both sides from t = to t = x, we have: x {P B (f(x))} = {P B (f(t))} dt = x P B (f(x)) P B (f()) = x. With the assumption that f() should correspond to the lowest number in the set B, we can safely set P B (f()) =. So, P B (f(x)) = x f(x) = P B (x). The CDF is an increasing function, so is its inverse. Moreover, P B () maps to the lowest number in the set B. So, it is a valid choice for f. In other words, to generate a point on the real line with probability distribution p B, simply apply the inverse of the CDF to a point x picked uniformly randomly from the interval [, )..3 Sampling from the Exponential Distribution dt We present a simple application of the inversion method. The exponential distribution with parameter λ is defined on [, ) with The CDF is given by: So, p(x) = λe λx. P (x) = e λx. P (y) = ln( y). Hence, to sample x acoording to the exponential distribution, we simply set: x := ln( ξ) where ξ is a randomly and uniformly sampled from the interval [, ).

3 3 Multi-Dimensional Case 3. The Main Theorem Let A, B R n, and p A ( ) be a probability density on A. Let f be given by: f (x, x,..., x n ) f (x, x,..., x n ) f(x, x,..., x n ) =. f n (x, x,..., x n ) be a function from A to B. The induced probability distribution p B arisen from the process of sampling a point x = (x, x,..., x n ) according to p A can then computing f(x) can again be computed by finding the volume of the image of the interval This volume is given by: [x, x + δx ) [x, x + δx ) [x n, x n + δx n ). Df(x) δx δx... δx n where f / x f / x f / x n f / x f / x f / x n Df(x) = f n / x f n / x f n / x n where all the partial derivatives are evaluated at x. Thus, p B (f(x)) = p A(x) Df(x). Notice that Df(x) is the factor that shows up when we perform change of variables during an integration. In two-dimensional space, we may write: [ ] x(u, v) f(u, v) =. y(u, v) In this case: Thus, Df(u, v) = x/ y/ x/ v y/ v. p A (u, v) p B (x, y) = x/ x/ v. y/ y/ v 3. The Polar Coordinate Transform The polar coordinate transforms two numbers (r, φ) to a point (x, y) on the plane as follows: x = r cos φ y = r sin φ, 3

4 which gives: x r = cos φ x = r sin φ φ y r = sin θ y φ = r cos φ So, if we sample a polar coordinate (r, φ) with probability distribution p A, then the distribution p B of the point (x, y) is given by: p A (r, φ) p p B (x, y) = cos φ A (r, φ) r sin φ = r cos φ + r sin φ = p A(r, φ). r sin θ r cos φ 3.3 Sampling Uniformly from the Unit Disk The unit disk is given by the polar coordinates in the set [, ] [, π). How should we be sampling the polar coordinates so that the resulting point distribution is uniform on the disk? In our case, we have that p B (x, y) = /π. So, we want p A such that: π = p A(r, φ) r p A (r, φ) = r π. A common strategy is to sample r and φ independently so that p A (r, φ) = p r (r)p φ (φ). Moreover, we shall sample φ uniformly from the interval [, π) so that p φ (φ) = /(π). Thus, p r (r) = r. The above distribution can be sampled with the inversion method. The CDF is given by: The inverse CDF is then: P r (r) = r r dr = [r ] r = r. P r (t) = t. So, we can sample points uniformly from the unit disk by setting: r := ξ φ = πξ where ξ and ξ are two independent random samples chosen uniformly from the interval [, ). 3.4 Sampling Uniformly from a Triangle Suppose we have a triangle in a plane with point A = (x A, y A ), B = (x B, y B ), C = (x C, y C ). Let us assume further that (B A) (C A) is pointing in the positive z-direction so that: area(abc) = (B A) (C A) = x B x A x C x A y B y A y C y A. 4

5 We wish to find a transformation f that takes a point (u, v) uniformly and randomly picked from the rectangle [, ) so that the distribution of (x, y) = f(u, v) is uniform on the triangle ABC. In this setting, we have that p A (u, v) =, and p B (x, y) = /area(abc). In other words, area(abc) = Df(u, v) Df(u, v) = x B x A y B y A x C x A y C y A. One way to generate a point on a triangle is to generate barycentric coordinates (α, β, γ) such that α, β, γ and α + β + γ =. Then, we can get a point on the triangle by computing In other words, (x, y) = αa + βb + γc = ( β γ)a + βb + γc = A + (B A)β + (C A)γ. x = x A + (x B x A )β + (x C x A )γ y = y A + (y B y A )β + (y C y A )γ. Our task is to figure out what β and γ are as functions of u and v. We have that x = (x B x A ) β + (x C x A ) γ x v = (x B x A ) β v + (x C x A ) γ v y = (y B y A ) β + (y C y A ) γ y v = (y B y A ) β v + (y C y A ) γ v. So, the matrix Df(u, v) is given by: [ (xb x Df(u, v) = A ) β + (x C x A ) γ (x B x A ) β v + (x C x A ) γ ] v (y B y A ) β + (y C y A ) γ (y B y A ) β v + (y C y A ) γ v [ ] [ ] xb x = A x C x A β/ β/ v. y B y A y C y A γ/ γ/ v Thus, Df(u, v) = x B x A x C x A y B y A y C y A β/ β/ v γ/ γ/ v x B x A x C x A y B y A y C y A = x B x A x C x A y B y A y C y A β/ β/ v γ/ γ/ v = β/ β/ v γ/ γ/ v β γ v β γ v =. What should β and γ be as functions of u and v? We have the constraint that β + γ. This condition is satisfied if we let β = g(u)( v) γ = g(u)v 5

6 where g(u) is a function such that g(u). With this choice of β and γ, we have that = β γ v β γ v = [g (u)( v)]g(u) [ g(u)][g (u)v] = g(u)g (u). It remains to find the function g with makes the above equation holds: g dg du = g dg = du g dg = du g = u g = u. Hence, a uniform distribution of points on triangle ABC can be generated by computing: ( u( v) uv)a + u( v)b + uvc where (u, v) is randomly and uniformly sampled from the rectangle [, ). 4 Dealing with 3D Manifolds 4. The Main Theorem Suppose that we have a differentiable function f that maps a set A R to a surface B R 3. We shall write: f x (u, v) f(u, v) = f y (u, v). f z (u, v) Again, let p A be a probability distribution on A. Given point (u, v) A, consider the rectangle [u + δu) [v + δv), which has area δuδv. This rectangle has probability mass p A (u, v)δuδv in it. We have that: where (u, v) f(u, v) (u + δu, v) f(u + δu, v) = f(u, v) + f u (u, v)δu (u, v + δv) f(u, v + δv) = f(u, v) + f v (u, v)δv (u + δu, v + δv) f(u + δu, v + δv) = f(u, v) + f u (u, v)δu + f v (u, v)δv f u (u, v) = f v (u, v) = f x f y f z f x v f y v f z v (u, v) (u, v), and (u, v) (u, v) (u, v). (u, v) In other words, the rectangle gets mapped to a parallelogram with sides defined by the vector f u (u, v)δu and f v (u, v)δv. The area of this parallelogram is given by: f u (u, v)δu f u (u, v)δv = f u (u, v) f u (u, v) δuδv 6

7 (Since the notation is getting a little unwieldy, let us drop the (u, v) arguments from the function from now on.) To compute the cross product, we make use of the following identity: So, Define We have that: a b + (a b) = (a a)(b b). f u f v = (f u f u )(f v f v ) (f u f v ) f u f v = (f u f u )(f v f v ) (f u f v ). E(u, v) = f u (u, v) f u (u, v) F (u, v) = f u (u, v) f v (u, v) G(u, v) = f v (u, v) f v (u, v). area of parallelogram = f u f v = EG F. In differential geometry, E, F, and G are called the coefficients of the first fundamental form. As a result, we have that the induced probability distribution is given by: p B (f(u, v)) = p A(u, v)δuδv f u f v δuδv = p A(u, v) EG F. 4. The Spherical Coordinate Transform The spherical coordinate is the transformation from (θ, φ) (, π) [, π) to a point ω on a 3D sphere S with: sin θ cos φ ω = sin θ sin φ. cos θ We then have that: So, cos θ cos φ ω θ = cos θ sin φ, sin θ sin θ sin φ ω φ = sin θ cos φ. E = cos θ cos φ + cos θ sin φ + sin θ = cos θ + sin θ = F = cos θ cos φ sin θ sin φ + cos θ sin φ sin θ cos φ + = G = sin θ sin φ + sin θ cos φ = sin θ EG F = sin θ = sin θ. 7

8 The inducted probability distribution is given by: p B (ω(θ, φ)) = p A(θ, φ) sin θ. However, since θ (, θ), we have that sin θ >. So, we can write: 4.3 Uniformly Sampling a Sphere p B (ω(θ, φ)) = p A(θ, φ) sin θ. We will use the identity to construct a sampling algorithm to sample a point on the unit sphere uniformly. The idea is to pick a probability distribution p A on (θ, φ) (, π) [, π) such that the induced probability distribution p B is the constant distribution /(4π). In other words: In other words: 4π = p A(θ, φ) sin θ. p A (θ, φ) = sin θ 4π. A common strategy is to sample φ independenty from θ so that p A (θ, φ) = p θ (θ)p φ (φ). Moreover, let us sample φ uniformly from [, π) so that p φ (φ) = /(π). In other words, p θ (θ) π = sin θ 4π p θ (θ) = sin θ. We can sample p θ (θ) using the inversion method. The CDF of p Θ is given by: P θ (θ) = θ So, the inverse function is given by: sin θ dθ = [ cos θ ] θ = cos cos θ = cos θ. P θ (u) = cos ( u). In conclusion, we compute θ and φ as: θ := cos ( ξ ) φ := πξ where ξ, ξ are two independent random numbers sampled uniformly from the interval [, ). Notice, however, that if the end goal is to get a point ω, there is no need to compute θ because θ never appears directly in the expression for ω. More specifically, sin θ cos φ ( ξ ) cos φ ω = sin θ sin φ = ( ξ ) sin φ. cos θ ξ 8

9 4.4 Sampling a Cosine-Weighted Hemisphere In this section, we want to sample the z-positive unit hemisphere such that the probability density being proportional to cos θ at each point. In this case: cos θ C = p A(θ, φ) sin θ C cos θ sin θ = p A(θ, φ), where C is the constant such that cos θ C is a probability distribution on the sphere. Again, we sample θ and φ independently with φ being uniform in [, π). So, The CDF then is given by: P θ (θ) = π C θ To determine C, note that P θ (π/) =, so π C cos θ sin θ = p θ(θ). cos θ sin θ dθ = π C [ cos θ = π C ( cos π ) = π C. In other words, C = π, and P θ (θ) = cos θ. Hence, we can sample the cosine-weighted hemisphere by setting: 4.5 From Area to Solid Angle cos θ := ξ φ := πξ. ] θ = π C ( cos θ). When shading from an area light source, a way to sample the incoming light direction is to sample a point on the light source s surface with some probability density p A and then convert the vector from the shaded point to the sampled point to a unit vector ω. In this section, we find the relation between p A and the induced probability density. For simplicity, let us say that the shaded point is at the origin and lying on the xy-plane so that the normal is the z-axis. Let r = (r x, r y, r z ) denote the point on the light source. Let n be the normal at r, and let s and t be the basis of the tangent plane at r in such a way that (s, t, n) is an orthonormal basis. The tangent plane is the set {r + us + vt (u, v) R }. The function f that maps the tangent plane to the direction is given by: ω = f(u, v) = Hence, using Lemma (proven in the appendix), we have: r + us + vt r + us + vt s f u (u, v) = r + us + vt r + us + vt (r s + u) r + us + vt 3 t f v (u, v) = r + us + vt r + us + vt (r t + v) r + us + vt 3 9

10 At (u, v) = (, ), we have that So, Next, f u (, ) = f v (, ) = s r r r 3 (r s) = s r r(r s) r 3 t r r r 3 (r t) = t r r(r t) r 3 E = r 4 r (r s) + r (r s) r 6 = r 4 r (r s) r 6 = r (r s) r 4 F = r (r s)(r t) (r s)(r t) r 6 = r 4 G = r (r t) r 4 EG F = r 4 r (r s) r (r t) + (r s) (r t) r 8 = r 4 r (r s) r (r t) r 8 = [ ( ) ( ) ] r r r 4 r s r t = r 4 [ (ˆr s) (ˆr t) ] (r s) (r t) r 8 where ˆr is the unit vector in the direction of r. Because s, t, n forms an orthonormal basis and ˆr =, we have that = ˆr = (ˆr s) + (ˆr t) + (ˆr n). So, Thus, EG F = r 4 [ (ˆr s) (ˆr t) ] = (ˆr n) r 4 EG F = (ˆr n) ˆr n r 4 = r. In conclusion, p B (f(r)) = r ˆr n p A(r) = r cos θ p A(r) 4.6 The Hair Coordinate System Transform The hair coordinate system maps (θ, φ) (π/, π/) [, π) to a sphere with the following transformation function: sin θ ω = cos θ cos φ. cos θ sin φ

11 So, cos θ ω θ = sin θ cos φ sin θ sin φ ω φ = cos θ sin φ cos θ cos φ E = cos θ + sin θ cos φ + sin θ sin φ = F = sin θ cos θ cos φ sin φ sin θ cos θ cos φ sin φ = G = cos θ sin φ + cos θ cos φ = cos θ EG F = cos θ = cos θ. However, since θ ( π/, π/), we have that cos θ >, so EG F = cos θ. So, the probability density transformation formula is: 4.7 Sampling for Diffuse Hair p B (ω(θ, φ)) = p A(θ, φ) cos θ. In this section, we want to sample the sphere so that p B (ω) cos θ. Applying the main theorem in this section, we have: cos θ C = p A(θ, φ) cos θ p A (θ, φ) = cos θ C. Again, we sample φ uniformly from [, π), and then sample θ independently from φ. So, To find C, we note that P θ (π/) =, so p θ (θ) = π C cos θ θ P θ (θ) = π cos θ dθ C π/ = π [ θ + sin θ cos θ ] θ C = π [ ] θ θ + sin(θ ) C π/ = π ( θ + sin(θ) + π ). C = π C ( π + + π ) = π C π/

12 So, C = π, and P θ (θ) = π ( θ + sin(θ) + π ). The above function cannot be inverted symbolically. So, in Mitsuba s implementation, they solve for it using Brent s method. 5 Appendix Lemma. Proof. a a = a a a ( a 3 a a ) a a = ( a a a ) a a = ( a a a ) a a a = ( a a a ( a a a )) a a = ( a a a a ( a a )) a = a a a ( a 3 a a )